a, Xét 1/2 < 2/3 ; 3/4<4/5 ; ............ ; 99/100<100/101

=> 1/2.3/4.......99/100 < 2/3.4/5.........100/101

=> M<N

b, M.N = 1/2.3/4.4/5......99/100.2/3.4/5.5/6......100/101

M.N = 1/2.2/3.3/4.4/5.............99/100.100/101

M.N = 1/101

c, Vì M<N nên M.M < M.N Hay M.M < 1/101 < 1/100

                                           hay M.M < 1/10 . 1/10

=> M < 1/10 (Đpcm)

 a) Ta có M.N = 1/2.2/3.3/4.4/5....99/10.10/101 = 1/101 
b) Xét M và N đều gồm 50 thừa số mà: 
1/2 < 2/3 
3/4 < 4/5 
............. 
99/100 < 100/101 
=> M < N 
c) Do M < N nên => M.M < M.N (Nhân 2 vế với M) 
=> M.M < 1/101 (Vì M.N = 1/101 theo cma) 
Mặt khác 1/101 < 1/100 
=> M.M < 1/100 = 1/10.1/10 
=> M < 1/10

Bài 2:

M = 1/2.3/4.5/6...99/100

Ta có: \(\frac{a}{b}\) = 1 - \(\frac{b-a}{b}\) (a; b; n ∈ N* và b > a)

\(\frac{a+n}{b+n}\) = 1 - \(\frac{b-a}{b+n}\)

\(\frac{a}{b}\) < \(\frac{a+n}{b+n}\)

Áp dụng công thức trên ta có:

\(\frac12<\frac{1+1}{2+1}=\frac23\)

\(\frac34<\frac{3+1}{4+1}=\frac45\)

\(\frac56\) < \(\frac{5+1}{6+1}\) = \(\frac67\)

............................

\(\frac{99}{100}\) < \(\frac{99+1}{100+1}\) = \(\frac{100}{101}\)

Cộng vế với vế ta có:

M = \(\frac12\).\(\frac34\).\(\frac56\)...\(\frac{99}{100}\) < \(\frac23\).\(\frac45\)..\(\frac{100}{101}\) = N

M < N (đpcm)

b; M.N = \(\frac12\).\(\frac34\).\(\frac56\)...\(\frac{99}{100}\).\(\frac23\).\(\frac45\)..\(\frac{100}{101}\)

M.N = \(\frac{1.3.5\ldots99}{3.5\ldots101}\). \(\frac{2.4.6\ldots100}{2.4.6\ldots100}\)

M.N = 1/100.101

\(M\cdot N=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{99}{100}\cdot\frac{100}{101}=\frac{1\cdot2\cdot3\cdot...\cdot\cdot\cdot.100}{2\cdot3\cdot4\cdot...\cdot100\cdot101}\)

\(=\frac{1}{101}\)

ta có           \(\frac{1}{2}< \frac{2}{3}\)

                   \(\frac{3}{4}< \frac{4}{5}\)

                    ................

                  \(\frac{99}{100}< \frac{100}{101}\)

NHÂN VẾ VỚI VẾ \(\Rightarrow M< N\)

c) \(M=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{1}{2}.\frac{4}{4}.\frac{6}{6}...\frac{100}{100}=\frac{1}{2}\)

a) M . N = \(\left(\frac{1}{2.}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)=\frac{1.2.3.4....100}{2.3.4.5...101}=\frac{1}{101}\)

Ta có:

M=\(\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}\)

M=\(\frac{1.3....99}{2.4....100}\)

Lại có:

N=\(\frac{2}{3}.\frac{4}{5}....\frac{100}{101}\)

N=\(\frac{2.4....100}{3.5....101}\)

\(\Rightarrow\)M.N=\(\frac{1.2.3......99.100}{2.3.4......100.101}\)

\(\Rightarrow\)M.N=\(\frac{1}{101}\)